Edhd 5007 expert_voices_rateand_work_problems_july26

3.
The members of a landscaping crew have the following individual times needed to mow a specific given yard: Jay: 6 hours Sally: 4 Rue: 3 Char: 8 Ming: 4.5 Q1a: What is the rate in terms of job/hr as the unit of speed? Use concept of distance = rate * time in terms of job or yard mowed = (amount of job per hour) * (time in hours) Jay rate = 1/6 job per hour or 1/6 job/hr Sally rate = 1/4 job/hr Rue rate = 1/3 job/hr Char rate = 1/8 job/hr Ming rate = 1/4.5 or 2/9 job/hr Q1b: For this yard, if the whole crew works together how long will it take them to complete? Answer is to add the total rates and find reciprocal (this uses the RT=D type formula mentioned in 1a)  1/6 + 1/4 + 1/3 + 1/8 + 2/9 using common denominator of 72 becomes (12 + 18 + 24 + 9 + 16) / 72 = 79/72 “job/hr” so take reciprocal is 72/79 hr per job i.e. just under 1 hour

4.
The members of a landscaping crew have the following individual times needed to mow a specific given yard: Jay: 6 hours Sally: 4 Rue: 3 Char: 8 Ming: 4.5 Q2: Which group would be faster? Jay, Char, and Ming vs. Sally and Rue? Just add the rates! (12 + 9 + 16)/72 job/hour vs. (18 + 24)/72 job/hour Team 1 total rate is  37/72 vs. Team 2 total rate is  42/72 The second group of Sally and Rue have the faster rate.

5.
The members of a landscaping crew have the following individual times needed to mow a specific given yard: Jay: 6 hours Sally: 4 Rue: 3 Char: 8 Ming: 4.5 Q3: Target rate for a new team member…Jay, Rue, and newcomer Cyrus are expected to finish this yard in 1 hour. What time would it take Cyrus alone to finish this yard in order for the team to meet the one hour target? The target of 1 job per 1 hour means that the rates must add to “1”. Cyrus rate is ‘1/x’. Summing the rates 1/6 + 1/3 + 1/x = 1/1 Common denominator of 6x yields, (x + 2x + 6)/6x = 1 3x + 6 = 6x “ algebra” obtains x = 2 as in 2hours for Cyrus to mow the yard alone. Check work 1/6 + 1/3 + 1/2 indeed = 1!

6.
The members of a landscaping crew have the following individual times needed to mow a specific given yard: Jay: 6 hours Sally: 4 Rue: 3 Char: 8 Ming: 4.5 Q4: Varying amounts of time with different team members. Key is to calculate portions of job per section of time. Jay and Sally work together on the same yard for 1 hour. Sally is called away, but is replaced by Char. Jay and Char continue to work together until this job or yard is completed Q4a : How much time do Jay and Char work together? Q4b (minor): What is total time to complete the job? Jay and Sally combined rate task. 1/6 + 1/4 yields 5/12 job per hour. After 1 hour 5/12 of job is done, 1 – (5/12) = 7/12 job remaining. Jay and Char combined rate task. 1/6 + 1/8 yields 7/24 job per hour. Rate * Time application!!! Use 7/24 * (TIME for Jay and Char) = 7/12. This obtains Time for Jay and Char of 2 hours for answer to 4a . 4b : 1 initial hour + 2 hours is trivial = 3 hours of total billable time!

7.
<ul><li>These word problems all lead or relate to rate * time = distance type formulas. Dimensional analysis from physical science helps in terms of familiarity. </li></ul><ul><li>Common denominator fractions and care with algebra are the computational tools. </li></ul><ul><li>There are other variations of these problems…Mowing could be painting... You could be filling a balloon with air or a tub with water… </li></ul><ul><li>A group or individual is working. A new person joins. A total time is given. Find rate of new person. (This is blend of Q3 and Q4) </li></ul><ul><li>Two people working (rates not given explicitly). Total time is given. Difference in individual times given. Ratio or quadratic likely solves problem i.e. find each rate. </li></ul><ul><li>A tub of water or balloon with air is being filled. Goal is to empty or drain. Something happens partway i.e. starts being filled. You have competing filling and emptying i.e. + rate and – rate depending on ‘relative’ definition of the rates. </li></ul>